Problem: $\overline{AC}$ is $10$ units long $\overline{BC}$ is $24$ units long $\overline{AB}$ is $26$ units long What is $\cot(\angle ABC)?$ $A$ $C$ $B$ $10$ $24$ $26$
Explanation: $\cot(\angle ABC) = \dfrac{1}{\tan(\angle ABC)}$ How can we find $\tan(\angle ABC)$ SOH CAH TOA angent = pposite over djacent Opposite $= \overline{AC} = 10$ Adjacent $= \overline{BC} = 24$ $\tan(\angle ABC) = \dfrac{10}{24}$ $\cot(\angle ABC) = \dfrac{1}{\tan(\angle ABC)} = \dfrac{24}{10}$